Mẹo What is the largest number that divides 437 732 and 1263 leaving a remainder of 24 in each case?
Kinh Nghiệm về What is the largest number that divides 437 732 and 1263 leaving a remainder of 24 in each case? 2022
Hoàng Hải Minh đang tìm kiếm từ khóa What is the largest number that divides 437 732 and 1263 leaving a remainder of 24 in each case? được Update vào lúc : 2022-10-03 05:56:04 . Với phương châm chia sẻ Kinh Nghiệm về trong nội dung bài viết một cách Chi Tiết 2022. Nếu sau khi tham khảo Post vẫn ko hiểu thì hoàn toàn có thể lại Comment ở cuối bài để Ad lý giải và hướng dẫn lại nha.Applications of the Fundamental Theorem of Arithmetic are finding the LCM and HCF of positive integers. So, this exercise đơn hàng with problems in finding the LCM and HCF by prime factorisation method. Also, the relationship between LCM and HCF is understood in the RD Sharma Solutions Class 10 Exercise 1.4. For quick access to complete solutions of this exercise, the RD Sharma Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.4 PDF is available below.
Nội dung chính- Access answers to Maths RD Sharma Solutions for Class 10 Chapter 1 Real Numbers Exercise 1.4What is the largest number that divides 967 and 1767 leaving remainder of 71 and 103 respectively?What is the greatest number which divides 42 and 142 leaving remainder 2 in each case?What is the largest number that divides 17220 and 420 leaving remainder 84 and 15 respectively?Which is the greatest number that divides 732 and 1174 leaving 3 and 4 respectively as remainders?
Access answers to Maths RD Sharma Solutions for Class 10 Chapter 1 Real Numbers Exercise 1.4
1. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = Product of the integers:
(i) 26 and 91
Solution:
Given integers are: 26 and 91
First, find the prime factors of 26 and 91.
26 = 2 × 13
91 = 7 × 13
∴ L.C.M (26, 91) = 2 × 7 × 13 = 182
And,
H.C.F (26, 91) = 13
Verification:
L.C.M × H.C.F = 182 x 13= 2366
And, product of the integers = 26 x 91 = 2366
∴ L.C.M × H.C.F = product of the integers
Hence verified.
(ii) 510 and 92
Solution:
Given integers are: 510 and 92
First, find the prime factors of 510 and 92.
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
∴ L.C.M (510, 92) = 2 × 2 × 3 × 5 × 23 × 17 = 23460
And,
H.C.F (510, 92) = 2
Verification:
L.C.M × H.C.F = 23460 x 2 = 46920
And, product of the integers = 510 x 92 = 46920
∴ L.C.M × H.C.F = product of the integers
Hence verified.
(iii) 336 and 54
Solution:
Given integers are: 336 and 54
First, find the prime factors of 336 and 54.
336 = 2 × 2 × 2 × 2 × 3 × 7
54 = 2 × 3 × 3 x 3
∴ L.C.M (336, 54) = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024
And,
H.C.F (336, 54) = 2 x 3 = 6
Verification:
L.C.M × H.C.F = 3024 x 6 = 18144
And, product of the integers = 336 x 54 = 18144
∴ L.C.M × H.C.F = product of the integers
Hence verified.
2. Find the LCM and HCF of the following integers by applying the prime factorization method:
(i) 12, 15 and 21
Solution:
First, find the prime factors of the given integers: 12, 15 and 21
For, 12 = 2 × 2 × 3
15 = 3 × 5
21 = 3 × 7
Now, L.C.M of 12, 15 and 21 = 2 × 2 × 3 × 5 × 7
∴ L.C.M (12, 15, 21) = 420
And, H.C.F (12, 15 and 21) = 3
(ii) 17, 23 and 29
Solution:
First, find the prime factors of the given integers: 17, 23 and 29
For, 17 = 1 × 17
23 = 1 × 23
29 = 1 × 29
Now, L.C.M of 17, 23 and 29 = 1 × 17 × 23 × 29
∴ L.C.M (17, 23, 29) = 11339
And, H.C.F (17, 23 and 29) = 1
(iii) 8, 9 and 25
Solution:
First, find the prime factors of the given integers: 8, 9 and 25
For, 8 = 2 × 2 x 2
9 = 3 × 3
25 = 5 × 5
Now, L.C.M of 8, 9 and 25 = 23 × 32 × 52
∴ L.C.M (8, 9, 25) = 1800
And, H.C.F (8, 9 and 25) = 1
(iv) 40, 36 and 126
Solution:
First, find the prime factors of the given integers: 40, 36 and 126
For, 40 = 2 x 2 x 2 × 5
36 = 2 x 2 x 3 x 3
126 = 2 × 3 × 3 × 7
Now, L.C.M of 40, 36 and 126 = 23 × 32 × 5 × 7
∴ L.C.M (40, 36, 126) = 2520
And, H.C.F (40, 36 and 126) = 2
(v) 84, 90 and 120
Solution:
First, find the prime factors of the given integers: 84, 90 and 120
For, 84 = 2 × 2 × 3 × 7
90 = 2 × 3 × 3 × 5
120 = 2 × 2 × 2 × 3 × 5
Now, L.C.M of 84, 90 and 120 = 23 × 32 × 5 × 7
∴ L.C.M (84, 90, 120) = 2520
And, H.C.F (84, 90 and 120) = 6
(vi) 24, 15 and 36
Solution:
First, find the prime factors of the given integers: 24, 15 and 36
For, 24 = 2 × 2 x 2 x 3
15 = 3 × 5
36 = 2 × 2 × 3 × 3
Now, LCM of 24, 15 and 36 = 2 × 2 × 2 × 3 × 3 × 5 = 23 x 32 x 5
∴ LCM (24, 15, 36) = 360
And, HCF (24, 15 and 36) = 3
3. Given that HCF (306, 657) = 9 , find LCM ( 306, 657 )
Solution:
Given two integers are: 306 and 657
We know that,
LCM × HCF = Product of the two integers
⇒ LCM = Product of the two integers / HCF
= (306 x 657) / 9 = 22338
4. Can two numbers have 16 as their HCF and 380 as their LCM? Give reason.
Solution:
On dividing 380 by 16 we get,
23 as the quotient and 12 as the remainder.
Now, since the LCM is not exactly divisible by the HCF its can be said that two numbers cannot have 16 as their HCF and 380 as their LCM.
5. The HCF of two numbers is 145 and their LCM is 2175. If one number is 725, find the other.
Solution:
The LCM and HCF of two numbers are 145 and 2175 respectively. (Given)
It is also given that, one of the numbers is 725
We know that,
LCM × HCF = first number × second number
2175 × 145 = 725 × second number
⇒ Second number = (2175 × 145)/ 725 = 435
∴ the other number is 435.
6. The HCF of two numbers is 16 and their product is 3072. Find their LCM.
Solution:
Given,
HCF of two numbers = 16
And, their product = 3072
We know that,
LCM × HCF = Product of the two numbers
LCM × 16 = 3072
⇒ LCM = 3072/ 16 = 192
∴ the LCM of the two numbers is 192.
7. The LCM and HCF of two numbers are 180 and 6 respectively. If one of the numbers is 30, find the other number.
Solution:
Given,
The LCM and HCF of two numbers are 180 and 6 respectively. (Given)
It is also given that, one of the numbers is 30.
We know that,
LCM × HCF = first number × second number
180 × 6 = 30 × second number
⇒ Second number = (180 × 6)/ 30 = 36
∴ the other number is 36.
8. Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.
Solution:
First let’s find the smallest number which is exactly divisible by both 520 and 468.
That is simply just the LCM of the two numbers.
By prime factorisation, we get
520 = 23 × 5 × 13
468 = 22 × 32 × 13
∴ LCM (520, 468) = 23 × 32 × 5 × 13 = 4680
Hence, 4680 is the smallest number which is exactly divisible by both 520 and 468 i.e. we will get a remainder of 0 in each case. But, we need to find the smallest number which when increased by 17 is exactly divided by 520 and 468.
So that is found by,
4680 – 17 = 4663
∴ 4663 should be the smallest number which when increased by 17 is exactly divisible by both 520 and 468.
9. Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.
Solution:
First, let’s find the smallest number which is exactly divisible by both 28 and 32.
Which is simply just the LCM of the two numbers.
By prime factorisation, we get
28 = 2 × 2 × 7
32 = 25
∴ L.C.M (28, 32) = 25 × 7 = 224
Hence, 224 is the smallest number which is exactly divisible by 28 and 32 i.e. we will get a remainder of 0 in each case. But, we need the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.
So that is found by,
224 – 8 – 12 = 204
∴ 204 should be the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.
10. What is the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in each case?
Solution:
First, let’s find the smallest number which is exactly divisible by all 35, 56 and 91.
Which is simply just the LCM of the three numbers.
By prime factorisation, we get
35 = 5 × 7
56 = 23 × 7
91 = 13 × 7
∴ L.C.M (35, 56 and 91) = 23 × 7 × 5 × 13 = 3640
Hence, 3640 is the smallest number that when divided by 35, 56 and 91 leaves the remainder of 7 in each case.
So that is found by,
3640 + 7 = 3647
∴ 3647 should be the smallest number that when divided by 35, 56 and 91 leaves the remainder of 7 in each case.
11. A rectangular courtyard is 18m 72cm long and 13m 20 cm broad. It is to be paved with square tiles of the same size. Find the least possible number of such tiles.
Solution:
Given,
Length of courtyard = 18 m 72 cm = 1800 cm + 72 cm = 1872 cm (∵1 m = 100 cm)
Breadth of courtyard = 13 m 20 cm = 1300 cm + 20 cm = 1320 cm
The size of the square tile needed to be paved on the rectangular yard is equal to the HCF of the length and breadth of the rectangular courtyard.
Now, finding the prime factors of 1872 and 1320, we have
1872 = 24 × 32 × 13
1320 = 23 × 3 × 5 × 11
⇒ HCF (1872 and 1320) = 23 × 3 = 24
∴ the length of side of the square tile should be 24 cm.
Thus, the number of tiles required = (area of the courtyard) / (area of a square tile)
We know that, area of the courtyard = Length × Breadth
= 1872 cm × 1320 cm
And, area of a square tile = (side)2 = (24cm)2
⇒ the number of tiles required = (1872 x 1320) / (24)2 = 4290
Thus, the least possible number of tiles required is 4290.
12. Find the greatest number of 6 digits exactly divisible by 24, 15 and 36.
Solution:
We know that, the greatest 6 digit number is 999999.
Let’s assume that 999999 is divisible by 24, 15 and 36 exactly.
Then, the LCM (24, 15 and 36) should also divide 999999 exactly.
Finding the prime factors of 24, 15, and 36, we get
24 = 2 × 2 × 2 × 3
15 = 3 × 5
36 = 2 × 2 × 3 × 3
⇒ L.C.M of 24, 15 and 36 = 360
Since, (999999)/ 360 = 2777 × 360 + 279
Here, the remainder is 279.
So, the greatest number which is divisible by all three should be = 999999 – 279 = 999720
∴ 999720 is the greatest 6 digit number which is exactly divisible by 24, 15 and 36.
13. Determine the number nearest to 110000 but greater 100000 which is exactly divisible by each of 8, 15 and 21.
Solution:
First, let’s find the L.C.M of 8, 15 and 21.
By prime factorisation, we have
8 = 2 × 2 × 2
15 = 3 × 5
21 = 3 × 7
⇒ L.C.M (8, 15 and 21) = 23 × 3 × 5 × 7 = 840
When 110000 is divided by 840, the remainder that is obtained is 800.
So, 110000 – 800 = 109200 should be divisible by each of 8, 15 and 21.
Also, we have 110000 + 40 = 110040 is also divisible by each of 8, 15 and 21.
⇒ 109200 and 110040 both are greater than 100000 but 110040 is greater than 110000.
Hence, 109200 is the number nearest to 110000 and greater than 100000 which is exactly divisible by each of 8, 15 and 21.
14. Find the least number that is divisible by all the numbers between 1 and 10 (both inclusive).
Solution:
From the question, it’s understood that
The LCM of 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 will be the least number that is divisible by all the numbers between 1 and 10.
Hence, the prime factors of all these numbers are:
1 = 1
2 = 2
3 = 3
4 = 2 × 2
5 = 5
6 = 2 × 3
7 = 7
8 = 2 × 2 × 2
9 = 3 × 3
10 = 2 × 5
⇒ L.C.M will be = 23 × 32 × 5 × 7 = 2520
Hence, 2520 is the least number that is divisible by all the numbers between 1 and 10 (both inclusive)
15. A circular field has a circumference of 360km. Three cyclists start together and can cycle 48, 60 and 72 km a day, round the field. When will they meet again?
Solution:
In order to calculate the time take before they meet again, we must first find out the individual time taken by each cyclist in covering the total distance.
Number of days a cyclist takes to cover the circular field = (Total distance of the circular field) / (distance covered in 1 day by a cyclist).
So, for the 1st cyclist, number of days = 360 / 48 = 7.5 which is = 180 hours [∵1 day = 24 hours]
2nd cyclist, number of days = 360 / 60 = 6 which is = 144 hours
3rd cyclist, number of days = 360 / 72 = 5 which is 120 hours
Now, by finding the LCM (180, 144 and 120) we’ll get to know after how many hours the three cyclists meet again.
By prime factorisation, we get
180 = 22 x 32 x 5
144 = 24 x 32
120 = 23 x 3 x 5
⇒ L.C.M (180, 144 and 120) = 24 x 32 x 5 = 720
So, this means that after 720 hours the three cyclists meet again.
⇒ 720 hours = 720 / 24 = 30 days [∵1 day = 24 hours]
Thus, all the three cyclists will meet again after 30 days.
16. In a morning walk three persons step off together, their steps measure 80 cm, 85 cm and 90 cm respectively. What is the minimum distance each should walk so that he can cover the distance in complete steps?
Solution:
From the question it’s understood that the required distance each should walk would be the L.C.M of the measures of their steps i.e. 80 cm, 85 cm, and 90 cm.
So, finding L.C.M (80, 85 and 90) by prime factorization we get,
80 = 24 × 5
85 = 17 × 5
90 = 2 × 3 × 3 × 5
⇒ L.C.M (80, 85 and 90) = 24 × 32 × 5 × 17 = 12240 cm = 122m 40cm [∵1 m = 100cm]
Hence, 122 m 40 cm is the minimum distance that each should walk so that all can cover the same distance in complete steps.
What is the largest number that divides 967 and 1767 leaving remainder of 71 and 103 respectively?
Step 1 - Decoding the information given in the question Remainder of 967x is 71. So, 967 – 71 = 896 is divisible by x. Remainder of 1767x is 103. So, 1767 – 103 = 1664 is divisible by x.What is the greatest number which divides 42 and 142 leaving remainder 2 in each case?
Solution: The GCF of 42 and 142 is 2.What is the largest number that divides 17220 and 420 leaving remainder 84 and 15 respectively?
Let x be the largest number that divides 170, 220, and 420 leaving reaminders of 8, 4, and 15 respectively. Remainder of 170x is 8. => 170 – 8 = 162 is divisible by x. Remainder of 220x is 4.Which is the greatest number that divides 732 and 1174 leaving 3 and 4 respectively as remainders?
Answer. The number will divide 729 without remainder. Tải thêm tài liệu liên quan đến nội dung bài viết What is the largest number that divides 437 732 and 1263 leaving a remainder of 24 in each case?
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